Do operator

The do operator is used in causal inference to denote an intervention. Given random variables $X,Y$, we write $Pr(Y=y\mid \mathit{d}\mathit{o}(X=x))$ to mean the probability that $Y=y$ given we intervene and set $X$ to be $x$. In some texts, this is abbreviated to $Pr(y\mid \hat{x})$ (this notation assumes that the random variables corresponding to the individual values are clear from context). The notation ${Pr}_x(y)$ is also used.

In general $Pr(Y=y\mid \mathit{d}\mathit{o}(X=x))$ is not the same as conditioning on $X=x$, i.e. $Pr(Y=y\mid X=x)$. Note also that in the expression $\mathit{d}\mathit{o}(X=x)$, the subexpression $X=x$ does not mean the event where the random variable $X$ takes on the value $x$, i.e. the event $\{\omega \in \mathrm{\Omega }:X(\omega )=x\}$. Thus, inside a do operator, the standard notational convention of probability theory does not hold. To stress the point, suppose the event $X=x$ can be specified in another way, such as by the event $Z=z$. In this case, since $X=x$ and $Z=z$ are exactly the same set, the probabilities involving them, such as $Pr(X=x)$ vs $Pr(Z=z)$ and $Pr(Y=y\mid X=x)$ vs $Pr(Y=y\mid Z=z)$, should all be the same, but I don't think $Pr(Y=y\mid \mathit{d}\mathit{o}(X=x))$ and $Pr(Y=y\mid \mathit{d}\mathit{o}(Z=z))$ need be the same (check this).

The do operator is used extensively in the do calculus.

History

Pearl: "An equivalent notation, using $\mathit{s}\mathit{e}\mathit{t}(x)$ instead of $\mathit{d}\mathit{o}(x)$, was used in Pearl (1995a). The $\mathit{d}\mathit{o}(x)$ notation was first used in Goldszmidt and Pearl (1992) and is gaining in popular support. Lauritzen (2001) used $P(y\mid X\leftarrow x)$. The expression $P(y\mid \mathit{d}\mathit{o}(x))$ is equivalent in intent to $P(Y_x=y)$ in the potential-outcome model introduced by Neyman (1923) and Rubin (1974) and to the expression $P[(X=x)◻\to (Y=y)]$ in the counter-factual theory of Lewis (1973b)."